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How to demonstrate that the $SU(4)$ group cannot be a group of symmetry of a color charge?

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Simplest way? The $\Delta^{++}\sim uuu$ has to be a color singlet. It has spin 3/2, so it is flavor and spin symmetric. But fermion quarks need to be in a fully antisymmetrized state. Can you make an SU(4) singlet out of three antisymmetrized copies of an SU(4) representation, the way you can for SU(3)? (No.)

(By now, you simply experimentally check R in $e^+e^-$, which may discriminate between 3 and 4.)

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  • $\begingroup$ And how to demonstrate it in a general way? $\endgroup$ Feb 23 at 17:12
  • $\begingroup$ By general way, you mean prove that no antisymmetric Kronecker product of any representation of SU(N) for 𝑁≠3, triplexed, can yield a singlet? An interesting group-theory exercise, but, by inspection of Slansky's tables, you see you'd need something truly screwed up and contrived, if possible at all. $\endgroup$ Feb 23 at 18:44

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